Question

Benzene reacts with CO/HCl/AlCl₃ and give 'P'.
Benzene reacts with CH₃-C-Cl / AlCl₃ to give 'Q'.
P and Q react with each other in presence of NaOH / H₂O give product Z. Find number of π e⁻ in product Z.

• A. 12
• B. 14
• C. 16
• D. 18

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question involves a multi-step organic synthesis starting from benzene.
First, we identify 'P' and 'Q' through Gatterman-Koch and Friedel-Crafts acylation reactions.
Then, we determine 'Z' through a crossed-aldol condensation between 'P' and 'Q'.
Step 2: Key Formula or Approach:
1. Gatterman-Koch Reaction: Benzene + $CO + HCl \xrightarrow{AlCl_3}$ Benzaldehyde ($C_6H_5CHO$).
2. Friedel-Crafts Acylation: Benzene + $CH_3COCl \xrightarrow{AlCl_3}$ Acetophenone ($C_6H_5COCH_3$).
3. Crossed Aldol Condensation: $C_6H_5CHO + C_6H_5COCH_3 \xrightarrow{NaOH/H_2O}$ Chalcone ($C_6H_5-CH=CH-CO-C_6H_5$).
Step 3: Detailed Explanation:
\includegraphics[width=0.5\linewidth]{16C_sol.png}
From the reaction sequence:
Product 'P' is Benzaldehyde.
Product 'Q' is Acetophenone.
When Benzaldehyde reacts with Acetophenone in base, the methyl group of acetophenone forms an enolate which attacks the carbonyl of benzaldehyde.
After dehydration, the product 'Z' is Chalcone: $C_6H_5-CH=CH-C(=O)-C_6H_5$.

Counting $\pi$ electrons in Z:
- Each Phenyl ring has 3 $\pi$ bonds $\times$ 2 $e^-$ = 6 $\pi$ electrons. Two rings = 12 $\pi$ electrons.
- The alkene ($C=C$) has 1 $\pi$ bond = 2 $\pi$ electrons.
- The carbonyl ($C=O$) has 1 $\pi$ bond = 2 $\pi$ electrons.
Total $\pi$ electrons = $12 + 2 + 2 = 16$.
Step 4: Final Answer:
The number of $\pi$ electrons in product Z is 16.